Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 79

Answer

$\color{blue}{\sqrt[3]{3}}$

Work Step by Step

Simplify each radical. Factor the radicand so that one factor is a perfect cube, then bring out the cube root of the perfect cube factor to obtain: $=2\sqrt[3]{3}+\sqrt[3]{8(3)} -\sqrt[3]{27(3)} \\=\sqrt[3]{3} + \sqrt[3]{2^3(3)}-\sqrt[3]{3^3(3)} \\=2\sqrt[3]{3}+2\sqrt[3]{3}-3\sqrt[3]{3}$ Combine like terms to obtain: $=(2+2-3)\sqrt[3]{3} \\=\color{blue}{\sqrt[3]{3}}$
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